# EQUATION OF TANGNET TO PARABOLA FROM EXTERNAL POINT

Equation of tangent drawn to the parabola will be in the form :

y = mx + (a/m)

Equation of tangent drawn to the ellipse will be in the form :

y = mx ± √a2 m2 + b2

Equation of tangent drawn to the hyperbola will be in the form

y = mx ± √a2 m2 + b2

## Point of Contact of Parabola Ellipse and Hyperbola

Point of contact of parabola :

Point of contact of Ellipse :

Point of contact of Hyperbola :

Problem 1 :

Find the equations of the tangents to the parabola y2 = 5x from the point (5, 13). Also find the points of contact.

Solution :

The equation of the parabola is y2 = 5x.

y2 = 4ax

5x = 4ax

a = 5/4

Equation of the tangent  y = mx + a/m

y = mx + 5/4m --- (1)

Passes through the points (5, 13).

13 = 5m + 5/4m

13 = (20m2 + 5)/4m

52m = 20m2 + 5

20m2 - 52m + 5 = 0

20m2 - 2m - 50m + 5 = 0

2m(10m - 1) - 5(10m - 1) = 0

(2m - 5) (10m - 1) = 0

2m - 5 = 0 and 10m - 1 = 0

2m = 5 and 10m = 1

m = 5/2 and m = 1/10

m = 5/2 substitute the equation (1).

m = 1/10 substitute the equation (1).

So, the points of contact are (1/5, 1), (125, 25).

Problem 2 :

Find the equation of the two tangents that can be drawn

From the point (2, -3) to the parabola y2 = 4x.

Solution :

(i)  Given, y2 = 4x

y2 = 4ax

4x = 4ax

a = 1

Passes through the points (2, -3).

Equation of the tangent to the parabola will be of the form y = mx + 1/m.

-3 = 2m + 1/m

-3 = (2m2 + 1)/m

-3m = 2m2 + 1

2m2 - 3m + 1 = 0

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